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Karnataka State Board · Maths

KARNATAKA STATE BOARD CLASS 10TH MATHS 2021-MCQ

2021 · 41 questions · First 10 Solutions Free
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Q1 mcq 1 mark
The nth term of an Arithmetic Progression is $a_n = 4n + 5$. Then its 5th term is
a.
20
b.
14
c.
25 Correct Answer
d.
24
Step-by-Step AI Explanation
Verified Method
The nth term is given by $a_n = 4n + 5$. To find the 5th term, substitute $n = 5$: $a_5 = 4(5) + 5 = 20 + 5 = 25$. Thus, option (c) is correct.
Q2 mcq 1 mark
When the quadratic equation $5x^2 = 2 ( 2x + 3 )$ is expressed in the standard form, the constant term obtained is
a.
5
b.
6
c.
4
d.
– 6 Correct Answer
Step-by-Step AI Explanation
Verified Method
Given the equation $5x^2 = 2(2x + 3)$, expand the right side: $5x^2 = 4x + 6$. Moving all terms to the left gives the standard form: $5x^2 - 4x - 6 = 0$. The constant term is $-6$. Thus, option (d) is correct.
Q3 mcq 1 mark
$x - 2y = 0$ and $3x + 4y - 20 = 0$ are
a.
Intersecting lines Correct Answer
b.
Coincident lines
c.
Parallel lines
d.
Perpendicular lines
Step-by-Step AI Explanation
Verified Method
For equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$, we compare the ratios: $\frac{a_1}{a_2} = \frac{1}{3}$ and $\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}$. Since $\frac{1}{3} \neq -\frac{1}{2}$, the lines are intersecting. Thus, option (a) is correct.
Q4 mcq 1 mark
The pair of equations of lines as shown in the graph are
Question 4 image
a.
x + y = 1 and 2x – y = 1
b.
2x + y = 2 and x + y = 2
c.
2x – y = 2 and 4x – y = 4 Correct Answer
d.
y – x = 0 and x – y = 1
Step-by-Step AI Explanation
Verified Method
From the graph, Line 1 passes through $(1, 0)$ and $(0, -2)$. Using the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, we get $\frac{x}{1} + \frac{y}{-2} = 1$, which simplifies to $2x - y = 2$. Line 2 passes through $(1, 0)$ and $(2, 4)$. Its slope is $\frac{4-0}{2-1} = 4$. Using $y - y_1 = m(x - x_1)$, $y - 0 = 4(x - 1) \implies y = 4x - 4 \implies 4x - y = 4$. Thus, option (c) is correct.
Q5 mcq 1 mark
If the pair of linear equations in two variables $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are parallel lines then the correct relation of their coefficients is
a.
\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} Correct Answer
b.
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}
c.
\frac{a_1}{a_2} \neq \frac{b_1}{b_2}
d.
\frac{a_1}{b_2} = \frac{b_1}{a_2}
Step-by-Step AI Explanation
Verified Method
A pair of linear equations represents parallel lines if and only if the ratios of the coefficients of $x$ and $y$ are equal, but the ratio of the constants is different, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$. Thus, option (a) is correct.
Q6 mcq 1 mark
If the pair of lines $2x + 3y + 7 = 0$ and $ax + by + 14 = 0$ are coincident lines then the values of ‘a’ and ‘b’ are respectively equal to
a.
2 and 3
b.
3 and 2
c.
4 and 6 Correct Answer
d.
1 and 2
Step-by-Step AI Explanation
Verified Method
For coincident lines, the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. Here, $\frac{2}{a} = \frac{3}{b} = \frac{7}{14} = \frac{1}{2}$. Thus, $\frac{2}{a} = \frac{1}{2} \implies a = 4$ and $\frac{3}{b} = \frac{1}{2} \implies b = 6$. Option (c) is correct.
Q7 mcq 1 mark
Which of the following is an Arithmetic Progression ?
a.
1, – 1, – 2, ..........
b.
1, 5, 9, .............. Correct Answer
c.
2, – 2, 2, – 2, ............
d.
1, 2, 4, 8, ..........
Step-by-Step AI Explanation
Verified Method
An Arithmetic Progression requires a constant common difference ($d = a_{n} - a_{n-1}$). For option (b): $5-1=4$, $9-5=4$. Since the difference is constant, it is an AP. The others do not have a constant difference.
Q8 mcq 1 mark
The 11th term of the Arithmetic Progression – 3, – 1, 1, 3, ....... is
a.
23
b.
– 23
c.
– 17
d.
17 Correct Answer
Step-by-Step AI Explanation
Verified Method
The general term of an AP is given by $a_n = a + (n-1)d$. Here, the first term $a = -3$ and the common difference $d = -1 - (-3) = 2$. For the 11th term ($n=11$): $a_{11} = -3 + (11-1)(2) = -3 + 10(2) = -3 + 20 = 17$. Option D is correct.
Q9 mcq
The sum of the first 10 terms of an Arithmetic Progression is 155 and the sum of the first 9 terms of the same progression is 126 then the 10th term of the progression is
a.
27
b.
126
c.
29 Correct Answer
d.
25
Step-by-Step AI Explanation
Verified Method
The $n^{th}$ term of an AP can be found using the formula $a_n = S_n - S_{n-1}$. Given $S_{10} = 155$ and $S_9 = 126$, the 10th term is $a_{10} = 155 - 126 = 29$. Option C is correct.
Q10 mcq
If one root of the equation $x^2 + ax + 6 = 0$ is 2, then the value of ‘a’ is
a.
7
b.
7/2
c.
– 7
d.
-7/2

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Q11 mcq
The discriminant of the Quadratic equation $px^2 + qx + r = 0$ is
a.
$q^2 - 4pr$
b.
$q^2 + 4pr$
c.
$p^2 - 4pr$
d.
$p^2 + 4qr$
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Q12 mcq
If 4, x, 10 are in Arithmetic Progression the value of x is
a.
14
b.
– 6
c.
– 7
d.
7
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Q12 short answer
Calculators, Mobiles, Smart Watches and any other electronic equipment are not allowed inside the examination hall.
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Q13 mcq
The roots of the quadratic equation $ax^2 + bx + c = 0$ are
a.
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
b.
$x = \frac{-b \pm \sqrt{b^2 + 4ac}}{2a}$
c.
$x = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$
d.
$x = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$
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Q14 mcq
The roots of the equation ( x – 3 ) ( x + 2 ) = 0 are
a.
– 3, 2
b.
3, – 2
c.
– 3, – 2
d.
3, 2
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Q15 mcq
If the sum of two consecutive integers is 27, then the integers are
a.
7 and 20
b.
13 and 14
c.
1 and 26
d.
– 13 and – 14
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Q16 mcq
In the figure, the value of sin  is
Question 17 image
a.
1/2
b.
√3/2
c.
√3
d.
2/√3
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Q17 mcq
The value of ( sin 30° + cos 60° – tan 45° ) is
a.
1
b.
– 1
c.
2
d.
0
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Q18 mcq
3 + sec2  is equal to
a.
4 + tan2 
b.
4 + cot2 
c.
2 + cot2 
d.
3 + cot2 
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Q19 mcq
The angle of elevation of the top of a tower from a point on the ground, which is 30 metres away from the foot of the tower, is 30°. Then the height of the tower is
a.
10 m
b.
30 m
c.
3√10 m
d.
30√3 m
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Q20 mcq
The value of ( sin   cosec  ) is
a.
2
b.
1
c.
– 1/2
d.
√3/2
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Q21 mcq
The formula to find the mid-point of the line segment joining the points A ( x1, y1 ) and B ( x2, y2 ) is
a.
( (x1+x2)/2, (y1+y2)/2 )
b.
( (x1-x2)/2, (y1-y2)/2 )
c.
( (x1+y1)/3, (x2+y2)/3 )
d.
( (x1+x2)/3, (y1+y2)/3 )
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Q22 mcq
The distance between the points ( $x_1, y_1$ ) and ( $x_2, y_2$ ) is
a.
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
b.
$\sqrt{(x_1 - x_2)^2 - (y_1 - y_2)^2}$
c.
$\sqrt{(x_2 + x_1)^2 + (y_2 - y_1)^2}$
d.
$\sqrt{(x_1 + x_2)^2 + (y_1 + y_2)^2}$
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Q23 mcq
The value among the observations of most repeated scores of the data is
a.
the mean
b.
the mode
c.
the median
d.
the range
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Q24 mcq
The Mean of the following scores is Marks: 1 3 5 7
a.
16
b.
5
c.
1.6
d.
4
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Q25 mcq
The relation among the Mean, Mode and Median is
a.
3 Median = 2 Mean + Mode
b.
3 Mean = 2 Median + Mode
c.
Mean = 3 Median + Mode
d.
Mode = 3 Mean + 2 Median
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Q26 mcq
A cylinder made of wax is melted and recast completely into a sphere. Then the volume of the sphere is
a.
two times the volume of the cylinder
b.
half the volume of the cylinder
c.
3 times the volume of the cylinder
d.
equal to the volume of the cylinder
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Q27 mcq
The formula to find the mid-point of the class interval is
a.
$\frac{\text{Upper limit} - \text{lower limit}}{2}$
b.
$\frac{\text{Upper limit} \times \text{lower limit}}{3}$
c.
$\frac{\text{Upper limit} + \text{lower limit}}{2}$
d.
$\frac{\text{Upper limit} + \text{lower limit}}{3}$
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Q28 mcq
In the $\Delta ABC$, $XY \parallel BC$ then
Question 29 image
a.
$\frac{AX}{AB} = \frac{AC}{AY}$
b.
$\frac{AX}{BX} = \frac{AY}{CY}$
c.
$\frac{AX}{BX} = \frac{XY}{AY}$
d.
$\frac{AB}{BX} = \frac{AC}{AY}$
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Q29 mcq
Observe the given two triangles and then identify the length of DF in the following :
Question 30 image
a.
26 cm
b.
23 cm
c.
4.2 cm
d.
8.4 cm
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Q30 mcq
$\Delta ABC \sim \Delta PQR$. Area of $\Delta ABC = 64 \text{ cm}^2$ and the area of $\Delta PQR = 100 \text{ cm}^2$. If $AB = 8 \text{ cm}$ then the length of $PQ$ is
a.
12 cm
b.
15 cm
c.
10 cm
d.
8 cm
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Q31 mcq
In the $\Delta ABC$, $\angle B = 90^\circ$ and $BD \perp AC$. If $AB = 6 \text{ cm}$, $BC = 8 \text{ cm}$ then the length of $CD$ is
Question 32 image
a.
10 cm
b.
6.4 cm
c.
4.8 cm
d.
3.6 cm
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Q32 mcq
In the given figure AT is a tangent drawn at the point A to the circle with centre O such that OT = 4 cm. If $\angle OTA = 30^\circ$ then AT is
Question 33 image
a.
4 cm
b.
2 cm
c.
$2\sqrt{3}$ cm
d.
$4\sqrt{3}$ cm
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Q33 mcq
In the given figure PA, PBC and CD are the tangents to a circle with centre O. If PC = 8 \text{ cm} and AP = 5 \text{ cm}, the length of the tangent CD is
Question 34 image
a.
5 \text{ cm}
b.
3 \text{ cm}
c.
8 \text{ cm}
d.
13 \text{ cm}
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Q34 mcq
The wrong statement in the following is
a.
(A) a tangent to a circle touches the circle exactly at one point
b.
(B) when a straight line is drawn to a circle it always passes through a point on the circle
c.
(C) the point common to the circle and its tangent is called the point of contact
d.
(D) the tangent drawn at any point to a circle is perpendicular to the radius drawn at the point of contact
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Q35 mcq
Which is the next step of construction while constructing a pair of tangents to a circle from an external point ‘T’, given in the figure ?
Question 36 image
a.
(A)
b.
(B)
c.
(C)
d.
(D)
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Q36 mcq
The surface area of a sphere is 616 sq.cm. Then the radius of the same sphere is
a.
49 cm
b.
14 cm
c.
21 cm
d.
7 cm
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Q37 mcq
The volume of a cone as shown in the figure is
Question 38 image
a.
\pi r^2 h
b.
\pi r ( r + l )
c.
\frac{1}{3} \pi r^2 h
d.
\pi rl
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Q38 mcq
The formula to find the total surface area of a right circular based cylindrical vessel of base radius r cm and height h cm opened at one end is
a.
(\pi r^2 + 2\pi rh) \text{ cm}^2
b.
2\pi rh \text{ cm}^2
c.
\frac{1}{3} \pi r^2 h \text{ cm}^3
d.
(\pi r^2 + h) \text{ cm}^2
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Q39 mcq
To find the curved surface area of a frustum of a cone as shown in the figure the formula used is
Question 40 image
a.
\frac{1}{3} \pi l (r_1 + r_2)
b.
\frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)
c.
\pi l (r_1 + r_2)
d.
\pi l (r_1 - r_2)
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Q40 mcq
The total surface area of solid hemisphere is 462 cm2 . If the curved surface area of it is 308 cm2 , then the area of the base of the hemisphere is
a.
308 \text{ cm}^2
b.
231 \text{ cm}^2
c.
154 \text{ cm}^2
d.
1078 \text{ cm}^2
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